Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{3x^2 - 15x}{x^2 - 100} \times \dfrac{x - 10}{x - 5} $
Solution: First factor the quadratic. $n = \dfrac{3x^2 - 15x}{(x - 10)(x + 10)} \times \dfrac{x - 10}{x - 5} $ Then factor out any other terms. $n = \dfrac{3x(x - 5)}{(x - 10)(x + 10)} \times \dfrac{x - 10}{x - 5} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ 3x(x - 5) \times (x - 10) } { (x - 10)(x + 10) \times (x - 5) } $ $n = \dfrac{ 3x(x - 5)(x - 10)}{ (x - 10)(x + 10)(x - 5)} $ Notice that $(x - 5)$ and $(x - 10)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 3x(x - 5)\cancel{(x - 10)}}{ \cancel{(x - 10)}(x + 10)(x - 5)} $ We are dividing by $x - 10$ , so $x - 10 \neq 0$ Therefore, $x \neq 10$ $n = \dfrac{ 3x\cancel{(x - 5)}\cancel{(x - 10)}}{ \cancel{(x - 10)}(x + 10)\cancel{(x - 5)}} $ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ $n = \dfrac{3x}{x + 10} ; \space x \neq 10 ; \space x \neq 5 $